Answer:
[tex]Q_m \simeq 0.11 \ m^3/sec[/tex]
Explanation:
From the given information:
The width of the prototype [tex]L_p[/tex] = 30 m
The flow of the prototype [tex]Q_p[/tex] = 100 m/s
width of the model [tex]L_m[/tex] = 2 m
The flow rate of the model can be calculated by using the geometrical similarities of Froude's Number Fr
[tex](Fr)_m = (Fr)_p[/tex]
[tex]\implies ( \dfrac{v}{\sqrt{ gy} }) _m = ( \dfrac{v}{\sqrt{gy}}) _p[/tex]
[tex]V_r = \sqrt{L_r}--- (1)[/tex]
[tex]Q_r = V_r *A_r[/tex]
[tex]Q_r = \sqrt { L_r} *L_r^2[/tex]
[tex]Q_r = L_r^{2.5}[/tex]
∴
[tex](\dfrac{Q_m}{Q_p}) = (\dfrac{L_m}{L_p})^{2.5}[/tex]
[tex](\dfrac{Q_m}{100}) = (\dfrac{2}{30})^{2.5}[/tex]
[tex](\dfrac{Q_m}{100}) =0.001147550621[/tex]
[tex]Q_m \simeq 0.11 \ m^3/sec[/tex]
